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Таблица истинности ONLINE
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Таблица истинности для функции F≡¬(C∨B)∧(A∨B):
Промежуточные таблицы истинности:C∨B: A∨B: ¬(C∨B): | C | B | C∨B | ¬(C∨B) | | 0 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 |
(¬(C∨B))∧(A∨B): | C | B | A | C∨B | ¬(C∨B) | A∨B | (¬(C∨B))∧(A∨B) | | 0 | 0 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 0 | 1 | 1 | 1 | | 0 | 1 | 0 | 1 | 0 | 1 | 0 | | 0 | 1 | 1 | 1 | 0 | 1 | 0 | | 1 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 0 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
F≡((¬(C∨B))∧(A∨B)): | F | C | B | A | C∨B | ¬(C∨B) | A∨B | (¬(C∨B))∧(A∨B) | F≡((¬(C∨B))∧(A∨B)) | | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
Общая таблица истинности:| F | C | B | A | C∨B | A∨B | ¬(C∨B) | (¬(C∨B))∧(A∨B) | F≡¬(C∨B)∧(A∨B) | | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности: | F | C | B | A | F | | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 |
F сднф = ¬F∧¬C∧¬B∧¬A ∨ ¬F∧¬C∧B∧¬A ∨ ¬F∧¬C∧B∧A ∨ ¬F∧C∧¬B∧¬A ∨ ¬F∧C∧¬B∧A ∨ ¬F∧C∧B∧¬A ∨ ¬F∧C∧B∧A ∨ F∧¬C∧¬B∧A Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности: | F | C | B | A | F | | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 |
F скнф = (F∨C∨B∨¬A) ∧ (¬F∨C∨B∨A) ∧ (¬F∨C∨¬B∨A) ∧ (¬F∨C∨¬B∨¬A) ∧ (¬F∨¬C∨B∨A) ∧ (¬F∨¬C∨B∨¬A) ∧ (¬F∨¬C∨¬B∨A) ∧ (¬F∨¬C∨¬B∨¬A) Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции | F | C | B | A | Fж | | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 1 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 0 | | 1 | 1 | 1 | 0 | 0 | | 1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина: F ж = C 0000 ⊕ C 1000∧F ⊕ C 0100∧C ⊕ C 0010∧B ⊕ C 0001∧A ⊕ C 1100∧F∧C ⊕ C 1010∧F∧B ⊕ C 1001∧F∧A ⊕ C 0110∧C∧B ⊕ C 0101∧C∧A ⊕ C 0011∧B∧A ⊕ C 1110∧F∧C∧B ⊕ C 1101∧F∧C∧A ⊕ C 1011∧F∧B∧A ⊕ C 0111∧C∧B∧A ⊕ C 1111∧F∧C∧B∧A Так как F ж(0000) = 1, то С 0000 = 1. Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы: F ж(1000) = С 0000 ⊕ С 1000 = 0 => С 1000 = 1 ⊕ 0 = 1 F ж(0100) = С 0000 ⊕ С 0100 = 1 => С 0100 = 1 ⊕ 1 = 0 F ж(0010) = С 0000 ⊕ С 0010 = 1 => С 0010 = 1 ⊕ 1 = 0 F ж(0001) = С 0000 ⊕ С 0001 = 0 => С 0001 = 1 ⊕ 0 = 1 F ж(1100) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 1100 = 0 => С 1100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0 F ж(1010) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 1010 = 0 => С 1010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0 F ж(1001) = С 0000 ⊕ С 1000 ⊕ С 0001 ⊕ С 1001 = 1 => С 1001 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0 F ж(0110) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0110 = 1 => С 0110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0 F ж(0101) = С 0000 ⊕ С 0100 ⊕ С 0001 ⊕ С 0101 = 1 => С 0101 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1 F ж(0011) = С 0000 ⊕ С 0010 ⊕ С 0001 ⊕ С 0011 = 1 => С 0011 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1 F ж(1110) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0010 ⊕ С 1100 ⊕ С 1010 ⊕ С 0110 ⊕ С 1110 = 0 => С 1110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0 F ж(1101) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0001 ⊕ С 1100 ⊕ С 1001 ⊕ С 0101 ⊕ С 1101 = 0 => С 1101 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 F ж(1011) = С 0000 ⊕ С 1000 ⊕ С 0010 ⊕ С 0001 ⊕ С 1010 ⊕ С 1001 ⊕ С 0011 ⊕ С 1011 = 0 => С 1011 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 F ж(0111) = С 0000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0001 ⊕ С 0110 ⊕ С 0101 ⊕ С 0011 ⊕ С 0111 = 1 => С 0111 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1 F ж(1111) = С 0000 ⊕ С 1000 ⊕ С 0100 ⊕ С 0010 ⊕ С 0001 ⊕ С 1100 ⊕ С 1010 ⊕ С 1001 ⊕ С 0110 ⊕ С 0101 ⊕ С 0011 ⊕ С 1110 ⊕ С 1101 ⊕ С 1011 ⊕ С 0111 ⊕ С 1111 = 0 => С 1111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0 Таким образом, полином Жегалкина будет равен: F ж = 1 ⊕ F ⊕ A ⊕ C∧A ⊕ B∧A ⊕ C∧B∧A Логическая схема, соответствующая полиному Жегалкина:
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